# Exponential of a differential operator

I have a differential operator $L$,

$\displaystyle L = i (t\frac{\partial}{\partial z} - z\frac{\partial}{\partial t})$

I can trivially hit this operator to $x,y,z$ and $t$ as $L x$, $L t$, $L y$, $L z$.

But I have a problem with exponential of that operator. I want to hit this operator to $x,y,z$ and $t$ as well.

$\exp(i\eta L)\,\,x$

($\eta$ is rapidity in this case)

The first thing that comes to my mind is to use definition of exponential of an operator:

$\displaystyle \exp(A) = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} ...$

But, I don't know why, I don't want to use this infinite sum. There should be a smart way of doing this..

Do you have any suggestions for me?

• Are you trying to obtain finite Lorentz boosts from the infinitesimal one? Mar 1, 2014 at 11:24
• I am trying to re-drive algebra structure of Lorentz Transformations with a different way. My operator actually is $J_{\mu\nu} = i(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu})$. First I am told that I need to deal with $J_{03}$, which L in this question. Mar 1, 2014 at 11:35

Suggestions:

1. Study how to derive the Lie algebra $so(p,q)$ from the Lie group $SO(p,q)$, cf. e.g. this Phys.SE post.

2. Work from now on at the level of Lie algebra (as opposed to Lie group). Show that $$\hat{J}^{\mu\nu}=\hat{x}^{\mu}\hat{p}^{\nu}-\hat{x}^{\nu}\hat{p}^{\mu}$$ are generators (of a representation) of the Lie algebra $so(p,q)$.

In this way the issue of Taylor series of the exponential map would be encountered only under pt. 1, and only in the form of finite-dimensional matrices (as opposed to differential operators).

I cannot understand why you wrote that $\eta$ is the rapidity: It would be the rapidity if $L$ were the boost, but this is not the case because the internal sign in the RHS of the formula defining $L$ is wrong. Your $L$ is formally an angular momentum if you do not pay attention to the weird name of the variable $t$, time?

Well, in addition to the procedures suggested by Qmechanic there is a third heuristic quite "brute force" way to get, however rapidly, the expression of $e^{i\eta L}$ in absence of any notion of Lie group representation theory. You could simply: (a) use the "universal relation" $$\left(e^{a \frac{d}{dx}}\right) f(x) = f(x+a)\qquad (1)$$ which evidently holds true at least for real analytic functions just because it is nothing but the Taylor expansion! And (b) you can change variables.

Let us start from: $$t = r \sin \tau\:, \quad z = r \cos \tau\qquad (2)$$ so $$\frac{\partial z}{\partial \tau} = -t\:, \quad \frac{\partial t}{\partial \tau} = z$$ and thus $$L = i \left( - \frac{\partial z}{\partial \tau} \frac{\partial}{\partial z}- \frac{\partial t}{\partial \tau} \frac{\partial}{\partial t}\right) = - i \frac{\partial}{\partial \tau}\:.$$ If you have a function $\psi(t,z)$ you can re-define: $$\phi(\tau, r):= \psi(t(\tau, r), z(\tau, r))\:,$$ so that exploiting (1): $$\left(e^{i\eta L}\psi\right)(t,z)= \left(e^{\eta \partial_\tau}\phi\right)(\tau,r)= \phi(\tau +\eta,r) = \psi(t(\tau+\eta, r), z(\tau+\eta, r))\:,\quad (3)$$ Since, from (2): $$t(\tau+\eta, r)= t\cos \eta + r \sin\eta\:, \qquad z(\tau+\eta, r)= r\cos \eta - t \sin\eta$$ we conclude, from (3), that: $$\left(e^{i\eta L}\psi\right)(t,z) = \psi\left( t\cos \eta + r \sin\eta, r\cos \eta - t \sin\eta\right)$$ As expected from the fact that $L$ is, formally, an angular momentum operator. If the sign in the RHS of the definition of $L$ were $+$, the same procedure could be used replacing $\sin$ with $\sinh$ and $\cos$ for $\cosh$ everywhere, eventually obtaining: $$\left(e^{i\eta L}\psi\right)(t,z) = \psi\left( t\cosh \eta + r \sinh\eta, r\cosh \eta + t \sinh\eta\right)\:.$$

WARNING: All this discussion is completely heuristic without mathematical guarantee for the validity of the obtained results which are, however, correct under suitable hypotheses on the employed space of functions and the topologies used to compute the exponential. For instance the formal Taylor expansion of an exponential like $e^{iA}$ is generally and incorrect procedure, leading to false results, if $A$ is an unbounded operator in a Hilbert or Banach space, also adopting the strong operator topology.
• You are right.. It should be $\displaystyle L = i (t\frac{\partial}{\partial z} + z\frac{\partial}{\partial t})$. Mar 3, 2014 at 20:26